To solve that equation, we may either use the quadratic formula, or notice that we may factor the quadratic to

$$(2x - 1)(x + 1) = 0$$

and so we see that $x = 1/2$ and $x = -1$ are both solutions to the quadratic. It is important to check that these are indeed solutions to our equation, since we do have two rational functions and they may very well be undefined on these points. Luckily, they do not cause our denominators to be $0$ in either case, both of these roots are solutions to our original problem.

Note. To solve the problem, we assumed that there was indeed some $x$ such that they are equal. This is not necessarily rigorous, per se, but it is fine for our purpose here.